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Mass distribution and gravitational field 
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#1
Oct2711, 07:20 AM

P: 11

The field outside a body with spherical mass distribution is well known but for a non simmetrical body (or a almost simmetrical body like the earth) which is the approach?. Do you know any reference?
thank you 


#2
Oct2711, 08:41 AM

P: 346

I think you can apply Gauss law. You can draw a spherical surface with center of center of mass of the object. When the sphere includes all the masses, according to gauss law, it is the same. Since Gauss theorem ▽·g=4πGρ, and apply the law ∫▽·g dV=∫g·dσ=4πM, where M is the mass inside the sphere. Therefore on this occasion where it is sufficiently far from the center of the object, you can use Newton.



#3
Oct2711, 09:44 AM

P: 617

Newton's law of gravity in its familiar form gives the gravitational force between two point mass objects (or two uniform spheres which act as point objects). However, it can be cast in a form that handles nonspherical masses. Just treat a nonspherical mass as a collection of integration volumes, each behaving as a point mass, and integrate over the volume.
Newton's Law in full fixedcoordinate vector form: F=  G m m' (x  x')/x  x'^{3} where F is the gravitational force of mass m' on m located at x' and x respectively. If another massive body m'' is introduced, it's force on m is just additive: F=  G m m' (x  x')/x  x'^{3}  G m m'' (x  x'')/x  x''^{3} If we bring many point masses into the picture, we just add up all the forces: F=  G m Ʃ m_{i}(x  x_{i})/x  x_{i}^{3} In the limit that the point masses get so packed together that they can be treated as constituting a spatial continuum of mass, the sum becomes a volume integral: F=  G m ∫ ρ(x')(x  x')/x  x'^{3} d^{(3)} x' where ρ is the mass density. This is what you would use for extended nonspherical mass objects. 


#4
Oct2711, 01:43 PM

P: 464

Mass distribution and gravitational field
You can also use Poisson's equation for the gravitational potential.



#5
Oct2911, 10:08 AM

P: 11

Thank you,
questions: what effects on the general motion of the earth are due to its nonspherical symmetry? astrophysicists will take account of this.... Even artificial satellites are definitely not spherically symmetric objects (although relatively small): as we can account for this? 


#6
Oct2911, 11:33 AM

P: 1




#7
Dec3112, 10:22 AM

P: 11

I return to the topic after a long time ...
A tangible example, the moon will not be a celestial body in perfect spherical symmetry. How this affect on its motion around the earth? thanks 


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